Let $H$ be a subspace of $V$.

For $c\in V,$ define $E(c) = \{c + h\,|\,h\in H\}$

Let $Q = \{ E(v)\,|\,v\in V\}$.

Define addition in $Q$ by: for $v, w\in V$, $E(v) \bigoplus E(w) = E(v+w)$

Define scalar multiplication in $Q$ by: for $v\in V$ and $\propto \in \mathbb{R}, \propto E(v) = E(\propto v)$.

Show that $Q$ together with this addition and scalar multiplication is a vector space.

My Attempt:

To check that $Q$ is a vector space, one must check each of the 10 axioms of a vector space to see if they hold.

$A_1:$ Let $a, b\in Q$. Then \begin{align*} a + b = E(a) + E(b) \in Q \end{align*} Therefore $Q$ is closed under addition ($A_1$ holds).

I haven't completed the whole solution yet, but you could imagine how long it would take. Is there a shorter and better way of proving?

  • $\begingroup$ I don't think there is a shorter way. On the other hand, I think all of the axioms are similarly short. It's not really that much work. $\endgroup$
    – Arthur
    Jun 2, 2021 at 11:16

1 Answer 1


You can see $Q$ as the subspace of the vector space $V$: Using your definition, we have $Q=\{v+H:v\in V\}$. This is nothing else than the quotient space (that's why the letter 'Q' is used :D) $Q=V/H$ with equivalence classes $[v]=v+H=\{v+h:h\in H\}$.

Let's take the direct sum $V=H\oplus H^\perp$ , then the quotient space is naturally isomorphic to the orthogonal complement of $H$: $$Q=V/H\simeq H^\perp$$.

This shows already, that $Q$ is isomorphic to a subspace of $V$ and thus a subspace itself.

See https://en.wikipedia.org/wiki/Quotient_space_(linear_algebra) for more details about Quotients of vector spaces.

Edit: If you really want to calculate the vector space axioms you can use the fact that the $+$ and $\cdot$ operation are inherited from $V$ and thus you only need to verify that $Q$ is a subspace. For this you only need to verify two axioms

  1. $Q\neq \emptyset$
  2. $\forall [v],[w]\in Q,\lambda\in\mathbb{R}$ we have $[v]+\lambda [w]\in Q$

and that should be easy.

  • $\begingroup$ Thank you for the informative answer. Just want to ask how do you exactly verify the second axiom you presented? $\endgroup$
    – muw
    Jun 2, 2021 at 11:34
  • 1
    $\begingroup$ Instead of $E(v)$ i prefer writing $[v]$. We then have $[v]+\lambda [w] = [v] + [\lambda w] = [v+\lambda w]$ using the addition and multiplication you had defined above. So $[v+\lambda w]$ is also an equivalence class and thus in $Q$ as well $\endgroup$
    – LegNaiB
    Jun 2, 2021 at 11:35
  • $\begingroup$ I see. The notation confused me. Thanks! $\endgroup$
    – muw
    Jun 2, 2021 at 11:36

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